Question 1058258: The digits 2, 3, 4, 5, 6, and 7 are randomly arranged to form a three-digit number. (Digits are not repeated.) Find the probability that the number is even and greater than 700. (Answer must be an integer or a simplified fraction.)
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! The digits 2, 3, 4, 5, 6, and 7 are randomly arranged to form a
three-digit number. (Digits are not repeated.) Find the
probability that the number is even and greater than 700.
(Answer must be an integer or a simplified fraction.)
We find the number of even three-digit numbers greater than
800 with all different digits taken from 2,3,4,5,6,7
The last digit is the MOST RESTRICTIVE, since it must be even.
so we choose it first:
We can choose the last digit any of 3 ways, as 2, 4, or 6.
We can then choose the first digit 1 way, as 7.
We can then choose the middle digit as any of the remaining 4 digits.
So that's 3∙1∙4 = 12 ways.
724, 726, 732, 734, 736, 742, 746, 752, 754, 756, 762, 764
So the numerator of the desired probability is 12.
The denominator is the number of permutations of 6 digits taken
3 at a time or 6P3 = 120. Or do it this way:
Choose the 1st digit 6 ways, the 2nd digit 5 ways, the 3rd digit 4 ways,
which is 6∙5∙4 = 120.
So the denominator of the desired probability is 120.
Answer is 12 ways out of 120 or 12/120 which reduces to 1/10.
Edwin
|
|
|
